2013 AIME II Problems/Problem 12
Let and
be real numbers so that the roots of
are complex conjugates. Enter the ordered pair
Two solutions of
are pure imaginary. Enter these solutions, separated by commas.
Solution 2 (Systematics)
This combinatorics problem involves counting, and casework is most appropriate. There are two cases: either all three roots are real, or one is real and there are two imaginary roots.
Case 1: Three roots are of the set . By stars and bars, there is
ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).
Case 2: One real root: one of . Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of
or
. Call the root
, where
is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of
tells us that we just need
to be integral, because
IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.)
Therefore, when the norm is
, the
term can range from
or
solutions. When the norm is
, the
term has
possibilities from
. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root,
, and you get
for this case.
And and we are done.
Solution 3 (Comments)
If the polynomial has one real root and two complex roots, then it can be factored as where
is real with
and
are integers with
The roots
and
are conjugates. We have
So
is either
or
. The only requirement for
is
All such quadratic equations are listed as follows:
where
where
.
Total of 130 equations, multiplied by 4 (the number of cases for real , we have 520 equations, as indicated in the solution.
-JZ
Solution 4
There are two cases: either all the roots are real, or one is real and two are imaginary.
Case 1: All roots are real.
Then each of the roots is a member of the set . It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same.
Sub-case 1.1: No two are the same.
This is obviously .
Sub-case 1.2: Exactly two are the same.
There are four ways to choose the root that will repeat twice, and three ways to choose the remaining root. For this sub-case, .
Sub-case 1.3: All three are the same.
This is obviously .
Thus for case one, we have polynomials in
. We now have case two, which we state below.
Case 2: Two roots are imaginary and one is real.
Let these roots be ,
, and
. Then by Vieta's formulas
;
;
.
Since ,
,
, and
are integers, we have that
for some integer
. Case two splits into two sub-cases now:
Sub-case 2.1: .
Obviously,
. The
cases in which
is either
are acceptable. Each can pair with one value of
and four values of
, adding
polynomials to
.
Sub-case 2.2: .
Obviously,
. Here, the
cases in which
is either
are acceptable. Again, each can pair with a single value of
as well as four values of
, adding
polynomials to
.
Thus for case two, polynomials are part of
.
All in all, polynomials can call
home.
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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